Expect an exponential

If $a=0$ then $P(n) = 0$ for all $n \ge 1$ , so that $f_n(X) \equiv 0$ for all $n \ge 1$ and hence $Pe(n) = 0$ for all $n \ge 1$ . The ratio $\frac{P(n) - Pe(n)}{P(n)}$ is not defined. If $a=1$ then $P(n) = 1$ for all $n \ge 1$ , so that $f_n(X) \equiv 1$ for all $n \ge 1$ , so that $Pe(n) = 1$ for all $n \ge 1$ , and hence $\frac{P(n) - Pe(n)}{P(n)} = 0$ for all $n \ge 1$ . Thus $b=0$ is one possible value.

Let us now assume that $a \neq 0,1$ . Lagrangian interpolation gives us that $\begin{aligned} f_n(X) & = \; \sum_{u=1}^n a^u \prod_{{1 \le v \le n} \atop {v \neq u}} \frac{X-v}{u-v} \; = \; \sum_{u=1}^n \frac{a^u}{(-1)^{n-u}(u-1)!(n-u)!}\prod_{{1 \le v \le n} \atop {v \neq u}} (X-v) \\ & = \; \frac{1}{(n-1)!} \sum_{u=1}^n (-1)^{n-u}\binom{n-1}{u-1}a^u \prod_{{1 \le v \le n} \atop {v \neq u}}(X-v) \end{aligned}$ and so $f_n(X)$ has degree $n-1$ and leading coefficient $\begin{aligned} \alpha_n & = \; \frac{1}{(n-1)!}\sum_{u=1}^n (-1)^{n-u}\binom{n-1}{u-1}a^u \; = \; \frac{1}{(n-1)!} \sum_{u=0}^{n-1} (-1)^{n-1-u}\binom{n-1}{u}a^{u+1} \\ & = \; \frac{1}{(n-1)!}a(a-1)^n \end{aligned}$ On the other hand we know that $f_{n+1}(X) - f_n(X)$ is a degree $n$ polynomial that vanishes at $1,2,...,n$ , and so it is clear that $f_{n+1}(X) - f_n(X) \; = \; \alpha_{n+1}(X-1)(X-2)\cdots (X-n)$ Putting $X = n+1$ we see that $\begin{aligned} f_{n+1}(n+1) - f_n(n+1) & = \; \alpha_{n+1} n! \\ P(n+1) - Pe(n+1) & = \; a(a-1)^n \end{aligned}$ so that $X_n \; = \; \frac{P(n+1) - Pe(n+1)}{P(n+1)} \; = \; \left(\frac{a-1}{a}\right)^n$ If $a > \tfrac12$ then $-1 < \tfrac{a-1}{a} < 1$ and hence $b=0$ . If $a=\tfrac12$ then $X_n = (-1)^n$ and so $b = \lim_{n \to\infty}X_n$ does not exist. If $0 < a < \tfrac12$ then $\tfrac{a-1}{a} < -1$ , and so the sequence $X_n$ fails to converge. If $a < 0$ then $\tfrac{a-1}{a} > 1$ and so the series $X_n$ fails to converge. Thus there is only $\boxed{1}$ possible value of $b$ , namely $b=0$ , which is achieved when $a > \tfrac12$ .